Where is the error in this 'proof' that the derivative of the quadratic function is equal to $x(2a+1)$? Note: I use $h$ to denote a small number.

$$f(x)=ax^2+bx+c\\f'(x)=\lim\limits_{h \to 0} \frac{a(x+h)^2+b(x+h)+c-(ax^2+bx+c)}{h}\\=\lim\limits_{h \to 0}\frac{ax^2+2ahx+ah^2+bx+hx+c-ax^2-bx-c}{h}\\=\lim\limits_{h \to 0}\frac{2ahx+ah^2+hx}{h}\\=\lim\limits_{h \to 0}2ax+ah+x\\=2ax+x\\=x(2a+1)$$

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    $\begingroup$ $bh$ became $hx$ on the third line. $\endgroup$ Dec 5 '19 at 22:16

You wrote $$\lim\limits_{h \to 0}\frac{ax^2+2ahx+ah^2+bx+\color{red}{hx}+c-ax^2-bx-c}{h}$$ instead of $$\lim\limits_{h \to 0}\frac{ax^2+2ahx+ah^2+bx+\color{blue}{bh}+c-ax^2-bx-c}{h}.$$

Simplifying, it becomes $$\lim\limits_{h \to 0}\frac{ax^2+2ahx+ah^2+bx+bh+c-ax^2-bx-c}{h}=\lim_{h\to0}\frac{h(2ax+b)}{h}=2ax+b$$ as desired.

  • $\begingroup$ Thank you. I'm glad I was only making a silly mistake. $\endgroup$
    – Joe
    Dec 5 '19 at 22:19

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